fameatop wrote:
Hi Bunuel,
Is there any other method (combinatorics) to solve this question ?
Following method may be applied-
As Bunuel has mentioned
"The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides."This a very important characteristic of triangle. [Thanks Bunuel, you are genius !!! ]
Now coming to this particular problem, we have sides mentioned as 10, 20, 30, 40, 50, 60.
Difference between any tow sides is 10, therefore 10 cannot be any side of a triangle.
So, we are left with 20, 30, 40, 50 and 60. ->> 5 sides.
Now, Number of triangles can be formed with 5 sides is \(^^5C {^}3\) \(= 10\)
However, following triangles are not possible-
\([ 20, 30, 50 ] -- as 20 + 30 (NOT) > 50\)
Similarly \([20, 30, 60 ]\) is not possible
Similarly \([20, 40, 60]\) is not possible
[Note: Easiest way to eliminate triangles is to start with least two sides and sum them up. All the sides having equal or higher that value are eliminated. In our case we started with 20 and 30. Hence, 50, and 60 are eliminated.
Repeat the same process with the next number along with the least one. In our case 20 and 40. So, 40 is eliminated.
]So, finally the number of triangles is \(10-3 =7\)
Hope, this is more logical and quicker method of solving such problem.